这个遍历方式也是LeetCode中 Binary Tree Inorder Traversal 一题的解法之一。 附题目,Binary Tree Inorder Traversal Given a binary tree, return theinordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[1,3,2]. Note:Recursive solution is trivial,...
Given a binary search tree, print the elements in-order iteratively without using recursion.Note:Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, coding a post-order iterative version is a ...
Preorder traversal starts printing from the root node and then goes into the left and right subtrees, respectively, while postorder traversal visits the root node in the end. #include<iostream>#include<vector>using std::cout;using std::endl;using std::string;using std::vector;structTreeNode{...
Output Specification: For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line. Sample Input: 10 1 2 3 4...
Preorder traversal Inorder traversal Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visi...
Preorder traversal Inorder traversal Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visi...
func(node*TreeNode)PostOrderTraversalPrint(){ifnode==nil{return}node.Left.PostOrderTraversalPrint()node.Right.PostOrderTraversalPrint()fmt.Println(node.Value)} 7.中序遍历 顺序是左子树、根节点、右子树。如下图所示 图片备用地址 trtree_in_orderee ...
*/classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right;}// if there is left, get the rightmost...
Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is...
publicList<Integer>inorderTraversal(TreeNoderoot){List<Integer>ans=newArrayList<>();getAns(root,ans);returnans;}privatevoidgetAns(TreeNodenode,List<Integer>ans){if(node==null){return;}getAns(node.left,ans);ans.add(node.val);getAns(node.right,ans);} ...